Question

Block 1, of mass m1 = 2.50 kg , moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 33.0 kg , which was initially at rest. The blocks stick together after the collision.A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.C. what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.

Answer 1

Answer:

a

The total initial momentum of the two-block system is  $p_t = 67.5 \ kg \cdot m/s^2$

b

The magnitude of the final velocity of the two-block system $v_f = 1.9014 \ m/s$

c

 the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is  

    $\Delta KE =- 847.08 \ J$

Explanation:

From the question we are told that

    The mass of first  block  is $m_1 = 2.50 \ kg$

      The initial velocity of first   block is $u_1 = 27.0 \ m/s$

          The mass of second block is  $m_2 = 33.0\ kg$

          initial velocity of second block is  $u_2 = 0 \ m/s$

         

The magnitude of the of the total initial momentum of the two-block system is mathematically repented as

        $p_i = (m_1 * u_1 ) + (m_2 * u_2)$

substituting values

        $p_i = (2.50* 27 ) + (33 * 0)$

        $p_t = 67.5 \ kg \cdot m/s^2$

According to the law of linear momentum conservation

        $p_i = p_f$

Where  $p_f$ is the total final momentum of the system which is mathematically represented as

       $p_f = (m_+m_2) * v_f$

Where $v_f$ is the final velocity of the system

      $p_i = (m_1 +m_2 ) v_f$

substituting values

       $67.5 = (2.50+33 ) v_f$

        $v_f = 1.9014 \ m/s$

The change in kinetic energy is mathematically represented as

     $\Delta KE = KE_f -KE_i$

Where $KE_f$ is the final kinetic energy of the two-body system  which is mathematically represented as

        $KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2$

substituting values

        $KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2$

        $KE_f =64.17 J$

While $KE_i$ is the initial kinetic energy of the two-body system

     $KE_i = \frac{1}{2} * m_1 * u_1^2$

substituting values

       $KE_i = \frac{1}{2} * 2.5 * 27^2$

        $KE_i = 911.25 \ J$

So

    $\Delta KE = 64.17 -911.25$

  $\Delta KE =- 847.08 \ J$